what is the speed with which a stone is projected vertically upwards from the ground if it attains a maximum height of 3.2m take g=10
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Answered by
5
Initial velocity=u. ,final velocity=0 and g=-10m/s^2 so by 3rd equation of motion, v^2=u^2+2as we get (0)^2 =u^2+2*(-10)(3.2). so we get. 0=u^2-64. u^2=64. Therefore u=8m/s
Answered by
14
heya friend
here is the answer of your question..
GIVEN:-
INITIAL VELOCITY ,u. = x
FINAL VELOCITY, v = 0
g = 10
height = 3.2 m
SOLUTION:-
ACC. TO third eqn of motion
v^2 = u^2- 2gh
0^2 = u^2-2×10×3.2
0= u^2 - 64
u^2= 64
u= √64
speed , u = 8m/s
... ....... ....... ....... ........ . . .
HOPE IT HELPS
PLS MARK ME AS BRAINLIEST...
:-)
here is the answer of your question..
GIVEN:-
INITIAL VELOCITY ,u. = x
FINAL VELOCITY, v = 0
g = 10
height = 3.2 m
SOLUTION:-
ACC. TO third eqn of motion
v^2 = u^2- 2gh
0^2 = u^2-2×10×3.2
0= u^2 - 64
u^2= 64
u= √64
speed , u = 8m/s
... ....... ....... ....... ........ . . .
HOPE IT HELPS
PLS MARK ME AS BRAINLIEST...
:-)
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